Question

Calculate the mL of 3.0 M NaOH can be added to a buffer composed of 0.015 mol CH3CO2H and 0.010 mol CH3CO2- in 500 mL of water before the buffer capacity is exceeded.

(The answer is 3.9 mL)

Answer 1

buffer capacity is defined as how much base can be added so the pH wont change more than 1 unit

First, find pH of buffer

let HA = CH3CO2H and A- = CH3CO2

use the buffer equation, Hendesron hasselbach equation:

for CH3CO2H , pKa = 4.75

pH = pKa + log(A-/HA)

substitute data

pH = 4.75 + log(0.010/0.015) = 4.5739

then, the max amount must exceed 4.5739 + 1 = 5.5739

substitute this in the pH equation

5.5739 = 4.75 + log(A-/HA)

note that Ha and A- will change as NaOH is added

HA + NaOH = H2O + A-

therefore, NaOH reacts with Ha to form more A-

mmol of A- = 10 + x

mmol of HA = 15 -x

where x is the mmol of NaOH added, let it be = Mbase*Vbase = 3*Vbase = x

mmol of A- = 10 + x = 10 + 3*Vbase

mmol of HA = 15 -x = 15- 3*Vbase

substitute in pH equation

5.5739 = 4.75 + log(A-/HA)

10^(5.5739-4.75) = (10 + 3*Vbase) / (15- 3*Vbase)

6.666 = (10 + 3*Vbase) / (15- 3*Vbase)

6.666*(15- 3*Vbase) = (10 + 3*Vbase)

99.99 - 19.998*Vbase = 10 + 3Vbase

(3+19.998)*Vbase = 99.99-10

Vbase = 89.99/((3+19.998)) = 3.91294 mL

therefore, we can't adde more thatn 3.9 mL, otherwise the buffer will exceed its capacity